ln(ln(ln(2x)))?

Question

How do you find the SECOND DERIVATIVE of this?

Sorry for the trouble but my prof really assigned this!!! Isn't is crazy? I got the 1st Derivative just fine but 2nd???

Answer 1

f(x) = ln(ln(ln(2x)))
f'(x) = 1 / (ln(ln(2x)) * 1/(ln(2x) * 2 / 2x
= 1 / ( x . ln(2x) . ln(ln(2x)) )
f''(x)= -1/( x² . ln(2x) . ln(ln(2x)) ) - 1/( x² . ln²(2x) . ln(ln(2x)) ) - 1/( x² . ln²(2x) . ln²(ln(2x)))

Answer 2

This is a real pain but it is also an extensive exercise in using the chain rule. It takes multiple substitutions, but with good bookkeeping, you can get the answer.

Your function is: y = ln(ln(ln(2*x)))

Do the substitutions:

v = ln(2*x)
w = ln(v) = ln(ln(2x))
y = ln(w) = ln(ln(ln(2x)))

The derivatives of these are:

dv/dx = 1/x
dw/dv = 1/v
dy/dw = 1/w

Start doing some chain rule applications to get the derivatives in terms of dx:

dw/dx = (dw/dv)*(dv/dx) = 1/(v*x)
y' = dy/dx = (dy/dw)*(dw/dv)*(dv/dx) = 1/(w*v*x)

Now with dy/dx, take the next derivative using the product rule and the chain rule:

y'' = d(dy/dx)/dx = -1/(w^2*v*x)*(dw/dx) – 1/(w*v^2*x)*(dv/dx) – 1/(w*v*x^2)*(dx/dx)

You should notice some pretty good symmetry here. Factor out a –1/(w*v*x) and put in the derivatives:

y'' = -1/(w*v*x)*((1/w)*1/(v*x) + (1/v)*(1/x) + (1/x)*1)

Simplify a little more:

y'' = -1/(w*v*x)*((1/w*v*x) + (1/v*x) + (1/x))

y'' = -(1 + w + w*v)/(w*v*x)^2

This is really not that bad until you substitute for w and v:

y'' = -(1 + ln(ln(2x)(1 + ln(2x)))/( ln(ln(2x))* ln(2x)*x)^2

Answer 3

By applying the chain rule a bunch of times. But... for crying out loud, who assigned this to you? This is sick if it is a high school assignment.

This is the answer: (thanks to Maple)

1 1 1
- ---------------------- - ----------------------- - ------------------------
2 2 2 2 2 2
x ln(2 x) ln(ln(2 x)) x ln(2 x) ln(ln(2 x)) x ln(2 x) ln(ln(2 x))

Too bad it doesn't post well... sorry.

Answer 4

Ooh --- what a fun question!!! Are you even serious?
I'm using D as d/dx throughout, okay?

D ln(2x) = 2*(1/2x)=1/x
So D ln(ln(2x)) = (1/x)(1/(ln(2x)))
So D ln(ln(ln(2x))) = (1/x)(1/ln(2x))/(ln(ln2x))=(1/(x*(ln(2x)*(ln(ln(2x))

D (1/ln(2x)) = -1/x/(ln(2x)^2
So.... D(1/(x(ln2x)) = -x^2/(ln(2x))+(1/x(-1/x(ln(2x)^2)=
-(1/x^2)(1/ln(2x))+1)(1/ln(2x))
D (1/ln(ln(2x))) = - (D ln(ln(2x)) / (ln(ln(2x)))^2

So DD ln(ln(ln(2x)))=
(-x^-2)(1+1/ln(2x))/ln(2x)*(1/ln(ln(2x)))+
1/(xln(2x))*(-(1/xln(2x)/ln(ln(2x))).

Well ---- that's the answer.... I'm beginning to think this was a cruel joke.

Answer 5

d of lnx = 1/x
f' = 1/x
d of 1/x = e^x
so... if x = ln(ln(ln(2x)))
f' = 1/ln(ln(2x)) ... you lost a ln()
f'' = e^(ln(ln(2x)) = f'' = ln(2x)... I think

Answer 6

Let y = ln u where u = lnv where v = ln(2x)

Then dy/dx = dy/du*du/dv*dv/dx
= 1/u * 1/v * 2/2x
= 1/ln(ln(2x)) *1/ln(2x) * 1/x =
______1______
x*ln(2x)*ln(ln(2x))

= (x*ln(2x)*ln(ln(2x)))^-1
= a^(-1)
d²y/dx² = d(a^(-1)) /da * da/dx = -1/a² * da/dx

da/dx =(d(x)/dx*ln(2x)*ln(ln(2x)) + x*d(ln(2x))/dx*ln(ln(2x))
+ x*ln(2x)*d(ln(ln(2x))/dx)
= 1*ln(2x)*ln(ln(2x)) + x*(1/x)*ln(ln(2x)) + x*ln(2x)*1/(xln(2x))
= ln2x*ln(ln(2x)) + ln(ln(2x)) + 1

So d²y/dx² =
-(ln2x*ln(ln(2x)) + ln(ln(2x)) + 1)
------------------- --------------------
......... (x*ln(2x)*ln(ln(2x)))²

WHEWWWWWWWWWWWWWWWWWW!!!!!!!!!! .... and I'd rather pass than do the third derivative (or should that be fail??? :-) )